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3 years ago

Draw a figure with the given perimeter. 6 units

1 answer:

6 0

We have to draw a figure with a perimeter 6 units. If we want to draw a rectangle it can`t be a 3 by 2 because it has a perimeter 2*2 + 2*3 = 4 + 6 = 10 units. It must be a 2 by 1 rectangle ( 2*2 + 2*1 = 4 + 2 = 6 units ).

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A building casts a shadow that is 50 feet in length. If the distance from the end of the shadow to the top of the building is 25

lozanna [386]

Answer:

The height of the building is of 244.95 feet.

Step-by-step explanation:

We use the Pythagorean Theorem to solve this question.

We have that:

The shadow of 50 feet is one side of the right triangle, while the height h is other side.

The hypotenuse is the distance from the end of the shadow to the top of the building, which is 250 feet.

$50^2 + h^2 = 250^2$

$h^2 = \sqrt{250^2 - 50^2}$

$h = 244.95$

The height of the building is of 244.95 feet.

8 0

2 years ago

Find an equivalent expression by simplifying 10x+5x-45+7y

egoroff_w [7]

Answer:

15x-45+7

Step-by-step explanation:

You would add the 10x and the 5x im pretty sure that would simplify it

3 0

3 years ago

Read 2 more answers

Please help me by showing how to do it

Margaret [11]

20 x 10 = 200
45 - 10 = 35 - 8 = 27
8 x 8 = 64
20 - 8 = 22
22 x 45 = 990
The area of the shape is 990 square feet.

Hope I could help!

8 0

3 years ago

Read 2 more answers

Square root of 1682/20000 and not a decimal plz

-Dominant- [34]

29/10000 is the square root of 1682/20000

4 0

3 years ago

Read 2 more answers

Suiting at 6 a.m., cars, buses, and motorcycles arrive at a highway loll booth according to independent Poisson processes. Cars

dem82 [27]

Answer:

Step-by-step explanation:

From the information given:

the rate of the cars = $\dfrac{1}{5} \ car / min = 0.2 \ car /min$

the rate of the buses = $\dfrac{1}{10} \ bus / min = 0.1 \ bus /min$

the rate of motorcycle = $\dfrac{1}{30} \ motorcycle / min = 0.0333 \ motorcycle /min$

The probability of any event at a given time t can be expressed as:

$P(event \ (x) \ in \ time \ (t)\ min) = \dfrac{e^{-rate \times t}\times (rate \times t)^x}{x!}$

∴

(a)

$P(2 \ car \ in \ 20 \ min) = \dfrac{e^{-0.20\times 20}\times (0.2 \times 20)^2}{2!}$

$P(2 \ car \ in \ 20 \ min) =0.1465$

$P ( 1 \ motorcycle \ in \ 20 \ min) = \dfrac{e^{-0.0333\times 20}\times (0.0333 \times 20)^1}{1!}$

$P ( 1 \ motorcycle \ in \ 20 \ min) = 0.3422$

$P ( 0 \ buses \ in \ 20 \ min) = \dfrac{e^{-0.1\times 20}\times (0.1 \times 20)^0}{0!}$

$P ( 0 \ buses \ in \ 20 \ min) = 0.1353$

Thus;

P(exactly 2 cars, 1 motorcycle in 20 minutes) = 0.1465 × 0.3422 × 0.1353

P(exactly 2 cars, 1 motorcycle in 20 minutes) = 0.0068

(b)

the rate of the total vehicles = 0.2 + 0.1 + 0.0333 = 0.3333

the rate of vehicles with exact change = rate of total vehicles × P(exact change)

$= 0.3333 \times \dfrac{1}{4}$

= 0.0833

∴

$P(zero \ exact \ change \ in \ 10 minutes) = \dfrac{e^{-0.0833\times 10}\times (0.0833 \times 10)^0}{0!}$

P(zero exact change in 10 minutes) = 0.4347

The probability of the 7th motorcycle after the arrival of the third motorcycle is:

$P( 4 \ motorcyles \ in \ 45 \ minutes) =\dfrac{e^{-0.0333\times 45}\times (0.0333 \times 45)^4}{4!}$

$P( 4 \ motorcyles \ in \ 45 \ minutes) =0.0469$

Thus; the probability of the 7th motorcycle after the arrival of the third one is = 0.0469

P(at least one other vehicle arrives between 3rd and 4th car arrival)

= 1 - P(no other vehicle arrives between 3rd and 4th car arrival)

The 3rd car arrives at 15 minutes

The 4th car arrives at 20 minutes

The interval between the two = 5 minutes

For Bus:

P(no other vehicle other vehicle arrives within 5 minutes is)

= $\dfrac{6}{12} = 0.5$

For motorcycle:

$= \dfrac{2 }{12} = \dfrac{1 }{6}$

∴

The required probability = $1 - \Bigg ( \dfrac{e^{-0.5 \times 0.5^0}}{0!} \times \dfrac{e^{-1/6}\times (1/6)^0}{0!} \Bigg)$

= 1- 0.5134

= 0.4866

6 0

3 years ago